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Laplace Transform
BIOE 4200
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Why use Laplace Transforms?
Find solution to differential equation
using algebra
Relationship to Fourier Transform
allows easy way to characterize
systems
No need for convolution of input and
differential equation solution
Useful with multiple processes in
system
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How to use Laplace
Find differential equations that describe
system
Obtain Laplace transform
Perform algebra to solve for output or
variable of interest
Apply inverse transform to find solution
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What are Laplace transforms?
F(s) L{f (t)} f (t)e dt
ò
ò
s+ ¥
f (t) L {F(s)} 1
1 st
s- ¥
= =
-
¥
-
p
= =
j
j
0
st
F(s)e ds
2 j
 t is real, s is complex!
 Inverse requires complex analysis to solve
 Note “transform”: f(t) ® F(s), where t is integrated
and s is variable
 Conversely F(s) ® f(t), t is variable and s is
integrated
Video.eAdshsoulem.ceosm f(t) = 0 for all t < 0

Evaluating F(s) = L{f(t)}
Hard Way – do the integral
let f (t) 1
¥
ò
=
(0 1) 1
s
= = - - =
¥ ¥
F(s) e e dt e dt 1
ò ò
at st (s a )t
0 0
f (t) sin t
¥
ò
- - - +
-
-
-
=
=
=
+
= = =
0
st
at
0
st
F(s) e sin(t)dt
s a
f (t) e
s
F(s) e dt 1
let
let
Integrate by parts
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Evaluating F(s)=L{f(t)}- Hard Way
remember òudv = uv - ò vdu
= - = - -
u e st ,du se stdt
= = -
dv sin(t)dt, v cos(t)
¥ ¥
e - sin(t)dt [ e - cos(t)] ¥ s e -
cos(t)dt
ò ò
= - - =
0 0
¥
ò
- -
st st
- -
0
st
0
st st
e (1) s e cos(t)dt
= - = - -
u e st ,du se stdt
= =
dv cos(t)dt, v sin(t)
=
ò ò
ò
¥
- -
¥
e cos(t)dt
]
- ¥ -
¥
-
st st
- + = - +
0
0
st
0
st
0
st
¥ ¥
ò ò
- -
st 2 st
= - =
se sin(t)dt 1 s e sin(t)dt
0 0
+ =
(1 s ) e sin(t)dt 1
[ e sin(t) s e sin(t)dt e (0) s e sin(t)dt
2
0
e st
sin(t)dt 1
2 st
0
+
1 s
=
ò
ò
¥
-
¥
-
let
let
Substituting, we get:
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Evaluating F(s) = L{f(t)}
This is the easy way ...
Recognize a few different transforms
See table 2.3 on page 42 in textbook
Or see handout ....
Learn a few different properties
Do a little math
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Table of selected Laplace
Transforms
f (t) = u(t) Û F(s) =
1
f (t) e u(t) F(s) 1
+
s a
-
= Û =
f (t) cos(t)u(t) F(s) s
2
s 1
f (t) sin(t)u(t) F(s) 1
s 1
s
2
at
+
= Û =
+
= Û =
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More transforms
f (t) t u(t) F(s) n!+ = Û =
n 1
n
s
120
n 0, f (t) u(t) F(s) 0!
= = Û = =
n = 1, f (t) = tu(t) Û F(s) =
1!
n 5,f (t) t u(t) F(s) 5!
1
6 6
5
1
2
s
s
s
s
s
= = Û = =
f (t) = d(t)ÛF(s) =1
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Note on step functions in Laplace
Unit step function definition:
= ³
u(t) 1, t 0
= <
u(t) 0, t 0
Used in conjunction with f(t) ® f(t)u(t)
because of Laplace integral limits:
¥
= -
ò
L{f (t)} f (t)e stdt
0
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Properties of Laplace Transforms
Linearity
Scaling in time
Time shift
“frequency” or s-plane shift
Multiplication by tn
Integration
Differentiation
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Properties: Linearity
L{c f (t) c f (t)} c F (s) c F (s) 1 1 2 2 1 1 2 2 + = +
Example :
- =
- =
) 1
s 1
L{sinh(t)}
e 1
2
t t
L{e } 1
2
t t
1
( 1
2
+ - -
((s 1) (s 1)
2
s 1
1
)
s 1
s 1
1
L{e }
2
1
e }
2
y{1
2 2
-
=
-
=
+
-
-
=
-
-
Proof :
+ =
L{c f (t) c f (t)}
1 1 2 2
+ =
[c f (t) c f (t)]e dt
ò ò
c f (t)e dt c f (t)e dt
c F (s) c F (s)
1 1 2 2
0
st
2 2
0
st
1 1
st
2 2
0
1 1
+
+ =
ò
¥
-
¥
-
-
¥
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Properties: Scaling in Time
F( s
a
)
a
L{f (at)} = 1
Example :
w Proof :
L{sin( t)}
1 ( 1
w
2 2
w
2 2
2
2
s
)
s
1 (
1)
( s )
+w
=
+w
w
+ =
w w
L{f (at)}
1
f (at)e dt
= = =
F( s
a
)
a
1
,dt 1
a
( s
f (u)e du
a
du
a
u at, t u
a
0
)u
a
0
st
ò
ò
¥
-
¥
-
=
=
=
let
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Properties: Time Shift
L{f (t - t )u(t - t )} = e -
st0F(s)
0 0
Example :
L{e u(t 10)}
e
10s
s a
a (t 10)
+
- =
-
- - Proof :
- - =
L{f (t t )u(t t )}
0 0
- - =
f (t t )u(t t )e dt
- =
f (t t )e dt
ò
u t t , t u t
0 0
- +
f (u)e du
st su
¥-
=
e 0 f (u)e du e 0
F(s)
0
0
0
st
0
t
0
s(u t )
t
st
0
0
st
0 0
-
¥
- -
¥
-
¥
-
ò
ò
ò
=
= - = +
let
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Properties: S-plane (frequency)
shift
L{e-atf (t)} = F(s + a)
Example :
- w = Proof :
L{e sin( t)}
w
2 2
at
+ +w
(s a)
at
-
L{e f (t)}
at st
e f (t)e dt
(s a )t
f (t)e dt
ò
0
¥
0
+
F(s a)
=
=
=
ò
- +
¥
- -
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Properties: Multiplication by tn
F(s)
n
L{t f (t)} ( 1) d n
ds
n = - n
Example :
L{t u(t)}
( 1) d
- =
n!
n 1
n
n
n
n
s
(1
)
s
ds
+
=
Proof :
n
= =
= - ¶
F(s)
s
n n st
L{t f (t)} t f (t)e dt
f (t)t e dt
n
- ¶
f (t)e dt ( 1)
n
- ¶
s
( 1)
e dt
s
( 1) f (t)
n
n
0
st
n
n
0
st
n
n
0
n st
0
¶
¶
=
¶
=
ò
ò
ò
ò
¥
-
¥
-
¥
-
¥
-
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The “D” Operator
1. Differentiation shorthand
2. Integration shorthand
Df (t) df (t)
f (t)
dt
=
D f (t) d
2
dt
2
2 =
t
if g(t) = ò
f (t)dt
= ò
-¥
then Dg(t) =
f (t)
then
t
g(t) f (t)dt
a
= -
1
a
g(t) D f (t)
if
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Properties: Integrals
L{D 1f (t)} F(s)
s
0 - =
Example :
- =
1
0
L{D cos(t)}
)( s
s
L{sin(t)}
) 1
s 1
s 1
(1
2 2
+
=
+
Proof :
let
¥
-
-
=
=
ò
L{sin(t)} g(t)e dt
0
st
1
0
= =
u g(t),du f (t)dt
dv e dt, v 1
st st
e
s
g(t) D f (t)
- -
= = -
f (t)e dt F(s)
s
[ 1
= - - ¥ + - =
ò
ò
=
t
0
st
0
st
g(t) f (t)dt
s
g(t)e ] 1
s
If t=0, g(t)=0
¥
ò
for (t = ¥ ) Þ f (t)e - stdt
< ¥
so
ò slower than
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e-st ®0
¥
= ®¥
0

Properties: Derivatives
(this is the big one)
L{Df (t)} = sF(s) - f (0+ )
Example :
L{Dcos(t)}
s
s
- =
+
- =
+
2 2
- +
s (s 1)
s 1
L{ sin(t)}
1
-
s 1
1
s 1
f (0 )
s 1
2
2
2
2
2
2
= -
+
+
=
+
Proof :
¥
L{Df (t)} d
f (t)e dt
dt
0
=
- -
= = -
u e ,du se
dv d
= =
f (t)dt, v f (t)
dt
[e f (t)] s f (t)e dt
0
f (0 ) sF(s)
st
0
st
st st
st
- +
+ =
+
¥
- ¥ -
-
ò
ò
let
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Difference in
f (0+ ), f (0- ) & f (0)
The values are only different if f(t) is not
continuous @ t=0
Example of discontinuous function: u(t)
= =
-
f (0 ) lim u(t) 0
®
t 0
-
= =
+
f (0 ) lim u(t) 1
®
t 0
+
= =
f (0) u(0) 1
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Properties: Nth order derivatives
L{D2f (t)} = ?
let = = =
g(t) Df (t),g(0) Df (0) f '(0)
2
= = - = -
L{D g(t)} sG(s) g(0) sL{Df (t)} f '(0)
2
= - - = - -
s(sF(s) f (0)) f '(0) s F(s) sF(0) f '(0)
L{Dnf (t)} = snF(s) - s(n-1)f (0) - s(n-2)f '(0) -- sf (n-2)' (0) - f (n-1)' (0)
NOTE: to take
you need the value @ t=0 for
called initial conditions!
L{Dnf (t)}
Dn-1f (t),Dn-2f (t),...Df (t),f (t)®
We will use this to solve differential equations!
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Properties: Nth order derivatives
Start with
Now apply again
L{Df (t)} = sF(s) - f (0)
L{D2f (t)}
g(t) = Df (t) and Dg(t) =
D2f (t)
= -
L{Dg(t)} sG(s) g(0)
=
g(t) Df (t)
=
g(0) f '(0)
= = = -
G(s) L{g(t)} L{Df (t)} sF(s) f (0)
L{Dg(t)} = sG(s) - g(0) = s[sF(s) - f (0)]- f '(0) = s2F(s) -sf (0) - f '(0)
D3f (t), D4f (t), etc.
let
then
remember
Can repeat for
L{Dnf (t)} = snF(s) - s(n-1)f (0) - s(n-2)f '(0) -- sf (n-2)' (0) - f (n-1)' (0)
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Relevant Book Sections
Modeling - 2.2
Linear Systems - 2.3, page 38 only
Laplace - 2.4
Transfer functions – 2.5 thru ex 2.4
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